# Missing Number

Updated: Mar 24, 2021

Given an array nums containing n distinct numbers in the range [0, n], return *the only number in the range that is missing from the array.*

**Follow up:** Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

**Example 1:**

**Input****:** nums = [3,0,1]
**Output****:** 2
**Explanation****:** n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

**Example 2:**

**Input****:** nums = [0,1]
**Output****:** 2
**Explanation****:** n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

**Example 3:**

**Input****:** nums = [9,6,4,2,3,5,7,0,1]
**Output****:** 8
**Explanation****:** n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.

**Example 4:**

**Input****:** nums = [0]
**Output****:** 1
**Explanation****:** n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.

**Constraints:**

n == nums.length

1 <= n <= 104

0 <= nums[i] <= n

All the numbers of nums are

**unique**.

**Solution:**

```
class Solution {
public int missingNumber(int[] nums) {
int sum1=0,sum2=0;
sum1=(nums.length*(nums.length+1))/2;
for(int i:nums) sum2+=i;
return sum1-sum2;
}
}
```